Dec. 1
Physics 207: Lecture 26, Pg 1
Lecture 26, Dec. 1
Goals:
• Chapter 19
 Understand the relationship between work and heat in a
cycling process
 Follow the physics of basic heat engines and refrigerators.
 Recognize some practical applications in real devices.
 Know the limits of efficiency in a heat engine.
• Assignment
 HW11, Due Friday, Dec. 5th
 HW12, Due Friday, Dec. 12th
 For Wednesday, Read through all of Chapter 20
Physics 207: Lecture 26, Pg 2
Heat Engines and Refrigerators
 Heat Engine: Device that transforms heat into work ( Q  W)
 It requires two energy reservoirs at different temperatures
 An thermal energy reservoir is a part of the environment so
large with respect to the system that its temperature doesn’t
change as the system exchanges heat with the reservoir.
 All heat engines and refrigerators operate between two
energy reservoirs at different temperatures TH and TC.
Physics 207: Lecture 26, Pg 3
Heat Engines
For practical reasons, we would like an engine to do the maximum
amount of work with the minimum amount of fuel. We can
measure the performance of a heat engine in terms of its thermal
efficiency η (lowercase Greek eta), defined as
We can also write the thermal efficiency as
Physics 207: Lecture 26, Pg 4
Exercise Efficiency
 Consider two heat engines:
 Engine I:
 Requires Qin = 100 J of heat added to system to
get W=10 J of work (done on world in cycle)
 Engine II:
 To get W=10 J of work, Qout = 100 J of heat is
exhausted to the environment
 Compare eI, the efficiency of engine I, to eII, the
efficiency of engine II.
Wcycle Qh  Qc
Qc


 1
Qh
Qh
Qh
Physics 207: Lecture 26, Pg 5
Exercise Efficiency
 Compare eI, the efficiency of engine I, to eII, the efficiency of
engine II.
 Engine I:
 Requires Qin = 100 J of heat added to system to get
W=10 J of work (done on world in cycle)
  = 10 / 100 = 0.10
 Engine II:
 To get W=10 J of work, Qout = 100 J of heat is exhausted
to the environment
 Qin = W+ Qout = 100 J + 10 J = 110 J
  = 10 / 110 = 0.09
Wcycle Qh  Qc
Qc


 1
Qh
Qh
Qh
Physics 207: Lecture 26, Pg 6
Refrigerator (Heat pump)
 Device that uses work to transfer heat from a colder object to a
hotter object.
K
QCold
WIn

What you get
What you pay
Physics 207: Lecture 26, Pg 7
The best thermal engine ever, the Carnot engine
 A perfectly reversible engine (a Carnot engine) can be
operated either as a heat engine or a refrigerator between the
same two energy reservoirs, by reversing the cycle and with no
other changes.
 A Carnot cycle for a gas engine
consists of two isothermal processes
and two adiabatic processes
 A Carnot engine has max. thermal
efficiency, compared with any other
engine operating between TH and TC
Carnot  1 
TCold
THot
 A Carnot refrigerator has a
maximum coefficient of performance,
compared with any other refrigerator
operating between TH and TC.
K Carnot 
TCold
THot TCold
Physics 207: Lecture 26, Pg 8
The Carnot Engine

Carnot showed that the thermal efficiency of a Carnot
engine is:
e Carnot cycle
Tcold
 1
Thot
 All real engines are less efficient than the Carnot engine
because they operate irreversibly due to the path and
friction as they complete a cycle in a brief time period.
Physics 207: Lecture 26, Pg 9
Problem
 You can vary the efficiency of a
Carnot engine by varying the
temperature of the cold
reservoir while maintaining the
hot reservoir at constant
temperature.
Which curve that best represents
the efficiency of such an
engine as a function of the
temperature of the cold
reservoir?
Temp of cold reservoir
Physics 207: Lecture 26, Pg 10
Other cyclic processes: Turbines
 A turbine is a mechanical device that extracts thermal energy
from pressurized steam or gas, and converts it into useful
mechanical work. 90% of the world electricity is produced by
steam turbines.
 Steam turbines &jet engines use a Brayton cycle
Physics 207: Lecture 26, Pg 13
Steam Turbine in Madison
 MG&E, the electric power plan in Madison, boils water to
produce high pressure steam at 400°C. The steam spins the
turbine as it expands, and the turbine spins the generator. The
steam is then condensed back to water in a Monona-lake-watercooled heat exchanger, down to 20°C.
 Carnot Efficiency?
Carnot  1 
TCold
THot
 1
 0.44
293 K
Physics673
207: Lecture
K 26, Pg 14
The Sterling Cycle
 Return of a 1800’s thermodynamic cycle
Isothermal
expansion
Isothermal
compression
SRS Solar System (~27% eff.)
Physics 207: Lecture 26, Pg 15
Sterling cycles
 1 Q, V constant  2 Isothermal expansion ( Won system < 0 ) 
1
3 Q, V constant  4 Q out, Isothermal compression ( Won sys> 0)
1 Q1 = nR CV (TH - TC)
2 Won2 = -nR TH ln (Vb / Va)= -Q2
3 Q3 = nR CV (TC - TH)
4 Won4 = -nR TL ln (Va / Vb)= -Q4
1
Gas
2
QCold = - (Q3 + Q4 )
T=TH
QHot = (Q1 + Q2 )
 = 1 – QCold / QHot
Gas
Gas
T=TH
T=TC
4
Gas
T=TC
3
P
1 2
x
4
start
Va
3
Vb
TH
TC
V
Physics 207: Lecture 26, Pg 16
Power from ocean thermal gradients… oceans
contain large amounts of energy
Carnot Cycle Efficiency
eCarnot = 1 - Qc/Qh = 1 - Tc/Th
See: http://www.nrel.gov/otec/what.html
Physics 207: Lecture 26, Pg 17
Ocean Conversion Efficiency
eCarnot = 1 - Tc/Th = 1 – 275 K/300 K
= 0.083 (even before internal losses
and assuming a REAL cycle)
Still: “This potential is estimated to be about 1013 watts of base load
power generation, according to some experts. The cold, deep seawater
used in the OTEC process is also rich in nutrients, and it can be used to
culture both marine organisms and plant life near the shore or on land.”
“Energy conversion efficiencies as high as 97% were achieved.”
See: http://www.nrel.gov/otec/what.html
So e =1-Qc/Qh is always correct but
eCarnot =1-Tc/Th only reflects a Carnot cycle
Physics 207: Lecture 26, Pg 18
Internal combustion engine: gasoline engine
 A gasoline engine utilizes the Otto cycle, in which fuel and air
are mixed before entering the combustion chamber and are
then ignited by a spark plug.
Otto Cycle
(Adiabats)
Physics 207: Lecture 26, Pg 19
Internal combustion engine: Diesel engine
 A Diesel engine uses compression ignition, a process by which
fuel is injected after the air is compressed in the combustion
chamber causing the fuel to self-ignite.
Physics 207: Lecture 26, Pg 20
Thermal cycle alternatives
 Fuel Cell Efficiency (from wikipedia)
Fuel cells do not operate on a thermal cycle. As such, they are not
constrained, as combustion engines are, in the same way by thermodynamic
limits, such as Carnot cycle efficiency. The laws of thermodynamics also hold
for chemical processes (Gibbs free energy) like fuel cells, but the maximum
theoretical efficiency is higher (83% efficient at 298K ) than the Otto cycle
thermal efficiency (60% for compression ratio of 10 and specific heat ratio of
1.4).
 Comparing limits imposed by thermodynamics is not a good predictor of
practically achievable efficiencies
 The tank-to-wheel efficiency of a fuel cell vehicle is about 45% at low loads
and shows average values of about 36%. The comparable value for a Diesel
vehicle is 22%.
 Honda Clarity
(now leased in CA and gets
~70 mpg equivalent)
This does not include H2
production & distribution
Physics 207: Lecture 26, Pg 21
Fuel Cell Structure
Physics 207: Lecture 26, Pg 22
Problem-Solving Strategy: Heat-Engine Problems
Physics 207: Lecture 26, Pg 23
Going full cycle
 1 mole of an ideal gas and PV= nRT  T = PV/nR
T1 = 8300 0.100 / 8.3 = 100 K
T2 = 24900 0.100 / 8.3 = 300 K
T3 = 24900 0.200 / 8.3 = 600 K T4 = 8300 0.200 / 8.3 = 200 K
(Wnet = 16600*0.100 = 1660 J)
12
P
DEth= 1.5 nR DT = 1.5x8.3x200 = 2490 J
2
3
24900
Wby=0
Qin=2490 J
QH=2490 J
N/m2
23
DEth= 1.5 nR DT = 1.5x8.3x300 = 3740 J
Wby=2490 J Qin=3740 J QH= 6230 J
4
1
8300
34
N/m2
DEth = 1.5 nR DT = -1.5x8.3x400 = -4980 J
Wby=0
Qin=-4980 J QC=4980 J
41
100
200 V
DEth = 1.5 nR DT = -1.5x8.3x100 = -1250 J
liters
liters
Wby=-830 J Qin=-1240 J QC= 2070 J
QH(total)= 8720 J QC(total)= 7060 J  =1660 / 8720 =0.19 (very low)
Physics 207: Lecture 26, Pg 24
Exercise
 If an engine operates at half of its theoretical maximum
efficiency (emax) and does work at the rate of W J/s, then, in
terms of these quantities, how much heat must be
discharged per second.
 This problem is about process (Q and W), specifically QC?
emax = 1- QC/QH and e = ½ emax = ½(1- QC/QH)
also W = e QH = ½ emax QH  2W / emax = QH
-QH (emax -1) = QC

QC = 2W / emax (1 - emax)
Physics 207: Lecture 26, Pg 25
Lecture 26, Dec. 1
• Assignment
 HW11, Due Friday, Dec. 5th
 HW12, Due Friday, Dec. 12th
 For Wednesday, Read through all of Chapter 20
Physics 207: Lecture 26, Pg 26